Answer:
A) An ionic compound dissolved in a polar solvent
Explanation:
Potassium Chloride's chemical formula is KCl. Most ionic compounds are formed between a nonmetal and a metal. In this case, potassium is acting as the metal and chloride is the nonmetal. Water is a polar solvent due to the electronegativity of the oxygen in the molecule creating a partial negative pole, leaving the hydrogen atoms partially positive. Hence, A is your best answer.
If I helped, a brainliest would be greatly appreciated!
Answer:
A: element B
B: element A
C: element B
D: element A
Explanation:
decrease in size leads increase in electronegativity because the smaller the size, the closer the shell is to the nucleus. Also, atomic radius decreases to the right and up on the periodic table. Atomic radius increases to the left and down a period. I hope this helps!
Carbon dating has<span> given archeologists a more accurate method by which they </span>can<span> determine the age of ancient artifacts. The </span>halflife<span> of </span>carbon 14<span> is </span>5730<span> ± 30 </span>years<span>, and the method of dating lies in trying to determine how </span>much carbon 14<span> (</span><span>the radioactive isotope of carbon) is present in the artifact and comparing it to levels</span>
Answer:
(a)
(b) 
Explanation:
The reaction that is carried out by the enzyme catalase produces
The reaction that is carried out by the enzyme catalase produces 
Answer:
M
Explanation:
Concentration of
= 0.020 M
Constructing an ICE table;we have:
![Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B4NH_3_%7Baq%7D%20%5Crightleftharpoons%20%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D_%7B%28aq%29%7D)
Initial (M) 0.020 0.40 0
Change (M) - x - 4 x x
Equilibrium (M) 0.020 -x 0.40 - 4 x x
Given that: 
![K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}](https://tex.z-dn.net/?f=K_f%20%7D%20%3D%20%5Cfrac%7B%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D%7D%7B%5BCu%5E%7B2%2B%7D%5D%5BNH_3%5D%5E4%7D)

Since x is so small; 0.40 -4x = 0.40
Then:








M